Lyapunov Matrix Equation Theorem REDUNDANT?
$begingroup$
I read the book A Linear Systems Primer by Antsaklis, Panos J., and Anthony N. Michel (Vol. 1. Boston: Birkhäuser, 2007) and I think a part of a theorem about the Lyapunov Matrix Equation seems wordy.
begin{equation}
dot{x}=Axtag{4.22}
end{equation}
Theorem 4.29. Assume that the matrix A [for system (4.22)] has no eigenvalues with real part equal to zero. If all the eigenvalues of A have negative real parts, or if at least one of the eigenvalues of A has a positive real part,then there exists a quadratic Lyapunov function
begin{equation}
v(x)=x^TPx,P=P^T
end{equation}
whose derivative along the solutions of (4.22) is definite (i.e., it is either negative definite or positive definite).
At the beginning of the theorem, it says "A has no eigenvalues with real part equal to zero" which means the eigenvalues of A either have negative real part or positive real part. So I think the sentence following this (marked bold in the theorem) is redundant and makes the theorem reads wordy. Am I right? Is it really necessary to write the bold sentence in this theorem?
matrix-equations stability-theory
asked Apr 17 '17 at 8:31
winstonwinston
524218
$endgroup$
add a comment |
$begingroup$
I read the book A Linear Systems Primer by Antsaklis, Panos J., and Anthony N. Michel (Vol. 1. Boston: Birkhäuser, 2007) and I think a part of a theorem about the Lyapunov Matrix Equation seems wordy.
begin{equation}
dot{x}=Axtag{4.22}
end{equation}
Theorem 4.29. Assume that the matrix A [for system (4.22)] has no eigenvalues with real part equal to zero. If all the eigenvalues of A have negative real parts, or if at least one of the eigenvalues of A has a positive real part,then there exists a quadratic Lyapunov function
begin{equation}
v(x)=x^TPx,P=P^T
end{equation}
whose derivative along the solutions of (4.22) is definite (i.e., it is either negative definite or positive definite).
At the beginning of the theorem, it says "A has no eigenvalues with real part equal to zero" which means the eigenvalues of A either have negative real part or positive real part. So I think the sentence following this (marked bold in the theorem) is redundant and makes the theorem reads wordy. Am I right? Is it really necessary to write the bold sentence in this theorem?
matrix-equations stability-theory
asked Apr 17 '17 at 8:31
winstonwinston
524218
$endgroup$
1
$begingroup$
I agree. Maybe the intention was to distinguish both cases, i.e. "the derivative is negative definite or positive definite respectively", although I am not sure this is true.
$endgroup$
– Miguel
Apr 18 '17 at 8:13
$begingroup$
Wordy does not mean the same thing as redundant. Wordiness is a matter of style, and asking if something makes some statement wordy is a slightly pointless question ;-) Asking if something is redundant, on the other hand, is a different matter!
$endgroup$
– Mariano Suárez-Álvarez
Apr 18 '17 at 16:26
$begingroup$
@MarianoSuárez-Álvarez OK. Maybe you are right.
$endgroup$
– winston
Apr 18 '17 at 23:56
add a comment |
$begingroup$
I read the book A Linear Systems Primer by Antsaklis, Panos J., and Anthony N. Michel (Vol. 1. Boston: Birkhäuser, 2007) and I think a part of a theorem about the Lyapunov Matrix Equation seems wordy.
begin{equation}
dot{x}=Axtag{4.22}
end{equation}
Theorem 4.29. Assume that the matrix A [for system (4.22)] has no eigenvalues with real part equal to zero. If all the eigenvalues of A have negative real parts, or if at least one of the eigenvalues of A has a positive real part,then there exists a quadratic Lyapunov function
begin{equation}
v(x)=x^TPx,P=P^T
end{equation}
whose derivative along the solutions of (4.22) is definite (i.e., it is either negative definite or positive definite).
At the beginning of the theorem, it says "A has no eigenvalues with real part equal to zero" which means the eigenvalues of A either have negative real part or positive real part. So I think the sentence following this (marked bold in the theorem) is redundant and makes the theorem reads wordy. Am I right? Is it really necessary to write the bold sentence in this theorem?
matrix-equations stability-theory
asked Apr 17 '17 at 8:31
winstonwinston
524218
$endgroup$
I read the book A Linear Systems Primer by Antsaklis, Panos J., and Anthony N. Michel (Vol. 1. Boston: Birkhäuser, 2007) and I think a part of a theorem about the Lyapunov Matrix Equation seems wordy.
begin{equation}
dot{x}=Axtag{4.22}
end{equation}
Theorem 4.29. Assume that the matrix A [for system (4.22)] has no eigenvalues with real part equal to zero. If all the eigenvalues of A have negative real parts, or if at least one of the eigenvalues of A has a positive real part,then there exists a quadratic Lyapunov function
begin{equation}
v(x)=x^TPx,P=P^T
end{equation}
whose derivative along the solutions of (4.22) is definite (i.e., it is either negative definite or positive definite).
At the beginning of the theorem, it says "A has no eigenvalues with real part equal to zero" which means the eigenvalues of A either have negative real part or positive real part. So I think the sentence following this (marked bold in the theorem) is redundant and makes the theorem reads wordy. Am I right? Is it really necessary to write the bold sentence in this theorem?
matrix-equations stability-theory
matrix-equations stability-theory
asked Apr 17 '17 at 8:31
winstonwinston
524218
asked Apr 17 '17 at 8:31
winstonwinston
524218
edited Apr 18 '17 at 23:57
winston
asked Apr 17 '17 at 8:31
winstonwinston
524218
asked Apr 17 '17 at 8:31
winstonwinston
524218
asked Apr 17 '17 at 8:31
winstonwinston
524218
524218
1
$begingroup$
I agree. Maybe the intention was to distinguish both cases, i.e. "the derivative is negative definite or positive definite respectively", although I am not sure this is true.
$endgroup$
– Miguel
Apr 18 '17 at 8:13
$begingroup$
Wordy does not mean the same thing as redundant. Wordiness is a matter of style, and asking if something makes some statement wordy is a slightly pointless question ;-) Asking if something is redundant, on the other hand, is a different matter!
$endgroup$
– Mariano Suárez-Álvarez
Apr 18 '17 at 16:26
$begingroup$
@MarianoSuárez-Álvarez OK. Maybe you are right.
$endgroup$
– winston
Apr 18 '17 at 23:56
add a comment |
1
$begingroup$
I agree. Maybe the intention was to distinguish both cases, i.e. "the derivative is negative definite or positive definite respectively", although I am not sure this is true.
$endgroup$
– Miguel
Apr 18 '17 at 8:13
$begingroup$
Wordy does not mean the same thing as redundant. Wordiness is a matter of style, and asking if something makes some statement wordy is a slightly pointless question ;-) Asking if something is redundant, on the other hand, is a different matter!
$endgroup$
– Mariano Suárez-Álvarez
Apr 18 '17 at 16:26
$begingroup$
@MarianoSuárez-Álvarez OK. Maybe you are right.
$endgroup$
– winston
Apr 18 '17 at 23:56
1
1
$begingroup$
I agree. Maybe the intention was to distinguish both cases, i.e. "the derivative is negative definite or positive definite respectively", although I am not sure this is true.
$endgroup$
– Miguel
Apr 18 '17 at 8:13
$begingroup$
I agree. Maybe the intention was to distinguish both cases, i.e. "the derivative is negative definite or positive definite respectively", although I am not sure this is true.
$endgroup$
– Miguel
Apr 18 '17 at 8:13
$begingroup$
Wordy does not mean the same thing as redundant. Wordiness is a matter of style, and asking if something makes some statement wordy is a slightly pointless question ;-) Asking if something is redundant, on the other hand, is a different matter!
$endgroup$
– Mariano Suárez-Álvarez
Apr 18 '17 at 16:26
$begingroup$
Wordy does not mean the same thing as redundant. Wordiness is a matter of style, and asking if something makes some statement wordy is a slightly pointless question ;-) Asking if something is redundant, on the other hand, is a different matter!
$endgroup$
– Mariano Suárez-Álvarez
Apr 18 '17 at 16:26
$begingroup$
@MarianoSuárez-Álvarez OK. Maybe you are right.
$endgroup$
– winston
Apr 18 '17 at 23:56
$begingroup$
@MarianoSuárez-Álvarez OK. Maybe you are right.
$endgroup$
– winston
Apr 18 '17 at 23:56
add a comment |
1 Answer
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It is only written for the detailed explanation. Nothing harms the clarity and accuracy of the whole statement.
answered Nov 30 '18 at 12:57
winstonwinston
524218
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$begingroup$
It is only written for the detailed explanation. Nothing harms the clarity and accuracy of the whole statement.
answered Nov 30 '18 at 12:57
winstonwinston
524218
$endgroup$
add a comment |
$begingroup$
It is only written for the detailed explanation. Nothing harms the clarity and accuracy of the whole statement.
answered Nov 30 '18 at 12:57
winstonwinston
524218
$endgroup$
add a comment |
$begingroup$
It is only written for the detailed explanation. Nothing harms the clarity and accuracy of the whole statement.
answered Nov 30 '18 at 12:57
winstonwinston
524218
$endgroup$
It is only written for the detailed explanation. Nothing harms the clarity and accuracy of the whole statement.
answered Nov 30 '18 at 12:57
winstonwinston
524218
answered Nov 30 '18 at 12:57
winstonwinston
524218
answered Nov 30 '18 at 12:57
winstonwinston
524218
answered Nov 30 '18 at 12:57
winstonwinston
524218
524218
add a comment |
add a comment |
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$begingroup$
I agree. Maybe the intention was to distinguish both cases, i.e. "the derivative is negative definite or positive definite respectively", although I am not sure this is true.
$endgroup$
– Miguel
Apr 18 '17 at 8:13
$begingroup$
Wordy does not mean the same thing as redundant. Wordiness is a matter of style, and asking if something makes some statement wordy is a slightly pointless question ;-) Asking if something is redundant, on the other hand, is a different matter!
$endgroup$
– Mariano Suárez-Álvarez
Apr 18 '17 at 16:26
$begingroup$
@MarianoSuárez-Álvarez OK. Maybe you are right.
$endgroup$
– winston
Apr 18 '17 at 23:56